Question:
If $f: R \rightarrow R$ be defined by $f(x)=\left(2-x^{5}\right)^{1 / 5}$, then $f \circ f(x)=$
Solution:
Given: $f(x)=\left(2-x^{5}\right)^{1 / 5}$
$f o f(x)=f(f(x))$
$=f\left(\left(2-x^{5}\right)^{1 / 6}\right)$
$=\left[2-\left(\left(2-x^{5}\right)^{1 / 5}\right)^{5}\right]^{1 / 5}$
$=\left[2-\left(2-x^{5}\right)^{\frac{1}{5} \times 5}\right]^{1 / 5}$
$=\left[2-2+x^{5}\right]^{1 / 5}$
$=\left[x^{5}\right]^{1 / 5}$
$=x^{5 \times 1 / 5}$
$=x$
Hence, if $f: R \rightarrow R$ be defined by $f(x)=\left(2-x^{5}\right)^{1 / 5}$, then $f o f(x)=\underline{x} .$
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